A sample of vinegar was found to have an acetic acid concentration of 0.8846 m. What is the acetic acid % by mass? Assume the density of vinegar is 1.00 g/ml.

5.3%

Explanation:

Let the volume be 1 L

volume , V = 1 L

use:

number of mol,

n = Molarity * Volume

= 0.8846*1

= 0.8846 mol

Molar mass of CH3COOH,

MM = 2*MM(C) + 4*MM(H) + 2*MM(O)

= 2*12.01 + 4*1.008 + 2*16.0

= 60.052 g/mol

use:

mass of CH3COOH,

m = number of mol * molar mass

= 0.8846 mol * 60.05 g/mol

= 53.12 g

volume of solution = 1 L = 1000 mL

density of solution = 1.00 g/mL

Use:

mass of solution = density * volume

= 1.00 g/mL * 1000 mL

= 1000 g

Now use:

mass % of acetic acid = mass of acetic acid * 100 / mass of solution

= 53.12 * 100 / 1000

= 5.312 %

≅ 5.3%

the partial pressure of methane CH₄ is 239.4 torr.

Explanation:

Given;

molar mass of methane, CH₄ = 90.0 g

molar mass of argon, Ar = 10.0 g

total pressure of the gases, Pt = 250 torr

number of moles of methane CH₄;

number of moles of Ar;

The mole fraction of methane XCH₄ is calculated as;

Apply Dalton's law of partial pressure, to determine the partial pressure of methane CH₄;

Therefore, the partial pressure of methane CH₄ is 239.4 torr.